CombatSystemMaintenanceStatusDavid H.OlwellIntroductionandsettingArmyforcesdependuponcombatsystemstomoveandfight.BecauseArmyforcesmoveinverystressfulenvironments,thesecombatsystemseventuallybreak and require repair.Whilea system is awaiting repair,it is considered nonmissioncapableorNMC.Thepercentage ofvehicles awaiting repairis one ofthe key indicators of a unit's ability to fight, and is tracked at all levels ofcommand.The operational readinessrate,or ORRate,isthepercentage ofsystemsinagivenclasswhichareMissionCapable(MC)In this paper,we will discuss three simplemodelswhichdescribetheORrateforthenumberoftanksinatankbattalion inthe1stArmoredDivision.WewillalsodiscusswaysthemodelscouldbeextendedtoimprovetheirusefulnesstothecommanderWeassumethatthereaderisfamiliarwithelementarymatrixalgebra,linearsystemsofdifferenceequations,and optimization using differential calculus.MarkovChainmodelsWebegin with a simple model where a tank can be inone of two states:missioncapable(MC)ornon-mission capable (NMC).Eachday,onaverageacertainpercentageof themission capabletanksbreak,andmovetotheother stateEachday,acertainpercentageoftheNMCtanksarerepairedandbecomeoperational.Graphically,wehave:169
169 Combat System Maintenance Status David H. Olwell Introduction and setting Army forces depend upon combat systems to move and fight. Because Army forces move in very stressful environments, these combat systems eventually break and require repair. While a system is awaiting repair, it is considered nonmission capable or NMC. The percentage of vehicles awaiting repair is one of the key indicators of a unit’s ability to fight, and is tracked at all levels of command. The operational readiness rate, or OR Rate, is the percentage of systems in a given class which are Mission Capable (MC). In this paper, we will discuss three simple models which describe the OR rate for the number of tanks in a tank battalion in the 1st Armored Division. We will also discuss ways the models could be extended to improve their usefulness to the commander. We assume that the reader is familiar with elementary matrix algebra, linear systems of difference equations, and optimization using differential calculus. Markov Chain models We begin with a simple model where a tank can be in one of two states: mission capable (MC) or non-mission capable (NMC). Each day, on average a certain percentage of the mission capable tanks break, and move to the other state. Each day, a certain percentage of the NMC tanks are repaired and become operational. Graphically, we have:
1-p11-p2DMCNMCp2Figure 1Weuse the notation that piis theprobability that a tank stays MC; 1-piis theprobabilitythatatank movesfromMCtoNMC;p2istheprobabilitythatatankmovesfromNMCtoMC,and1-P2istheprobabilitythatatankstaysNMC.Let M,bethenumberoftanks inthebattalion whichare mission capable ondayi, and N,bethe number oftanks non-mission capable.Then we can writetheexpectedtransitionsfromonedaytothenext:MuI= p,M, + p,N,Ni+1I = (1- PI) M, +(1- P2)N,This systemofequationscanbeeasilyexpressed inmatrixnotation:M.IMHPiP2HPnM1-p.1-p.U1'N..3This assumes thatthe p,areknownand constant,and ignores therandomnessinthisproblemby just lookingattheexpectedvalues oftanks operational.(Recall the expected value of an integer-valued variable is not necessarilyaninteger.)Thissystemwill eventuallyreachasteadystate,wheretheexpectedvaluesdonot change fromday to day.The steadystate can befoundby eigenvectoranalysis.A matrix ofprobabilities such as wehave constructed is called aprobability transition matrix, since every entry is non-negative and every column170
170 Figure 1 We use the notation that p1 is the probability that a tank stays MC ; 1-p1 is the probability that a tank moves from MC to NMC; p2 is the probability that a tank moves from NMC to MC, and 1- p2 is the probability that a tank stays NMC. Let Mi be the number of tanks in the battalion which are mission capable on day i, and Ni be the number of tanks non-mission capable. Then we can write the expected transitions from one day to the next: M p M p N N p M p N i i i i i i + + = + = - + - 1 1 2 1 1 1 1 2 ( ) ( ) This system of equations can be easily expressed in matrix notation: M N p p p p M N p p p p M N i i i i i + + + L N M O Q P = - - L N M O Q P L N M O Q P = - - L N M O Q P L N M O Q P 1 1 1 2 1 2 1 2 1 2 1 0 1 1 1 1 0 This assumes that the pi are known and constant, and ignores the randomness in this problem by just looking at the expected values of tanks operational. (Recall the expected value of an integer-valued variable is not necessarily an integer.) This system will eventually reach a steady state, where the expected values do not change from day to day. The steady state can be found by eigenvector analysis. A matrix of probabilities such as we have constructed is called a probability transition matrix, since every entry is non-negative and every column
sumsto 1.Itisknownthat everyprobabilitytransition matrixhas at leastoneeigenvalueequal to1,and itsassociated eigenvectoristhesteadystateforthesystem.Wecan solve forthe steady state eigenvector by solving the equationMPiP2p1-pNsinceat the steady statethenumber of vehiclesbreaking exactlybalancesthenumberofvehiclesbeingrepaired.WeobtaintheeigenvectorT p2p2-p +1pMT(-p)PNp,-p,+1bwhereT isthetotal number of tanks in thebattalion(T=M+N).Regardlessofourstartingstate,thetop componentoftheeigenvectortellsushowmanyvehicles,onaverage,wecan expecteventuallytohaveworking,thebottomcomponenttellsushowmanyweexpecttobeawaiting repair.TheMC/NMCratesteadystatescanbefoundsimilarlytobeP2P2-P,+1p(1-p)Np2 - p, +10For example, let's let p, =.95 and p, =.8. This means that only 5% of theworkingvehiclesbreakeachday,and 80%of thebrokenvehicles arerepaired164.58826Weset T=58.Substitutiongives us a steadystate ofandan3.4118eventual steadystateORof 94.1176%.Ofcourse,wecan'thave54.5882tanks.That is theexpected valuefor the number operational, whichis anaverage. Just as the average of 1 and 2 is 1.5, which is not an integer, so canthe averagenumber oftanks operational also notbean integer.We can verify these calculations by constructing a spreadsheet model of thetank maintenance status. Let's start with M = 50 and N =4. Figure 2 is a pictureofhowthesystemonaveragewouldbehave171
171 sums to 1. It is known that every probability transition matrix has at least one eigenvalue equal to 1, and its associated eigenvector is the steady state for the system. We can solve for the steady state eigenvector by solving the equation M N p p p p M N L N M O Q P = - - L N M O Q P L N M O Q P 1 2 1 1 1 2 since at the steady state the number of vehicles breaking exactly balances the number of vehicles being repaired. We obtain the eigenvector T p p p T p p p 2 2 1 1 2 1 1 1 1 - + - - + L N M M M M O Q P P P P ( ) , where T is the total number of tanks in the battalion (T = Mi Ni + ). Regardless of our starting state, the top component of the eigenvector tells us how many vehicles, on average, we can expect eventually to have working, the bottom component tells us how many we expect to be awaiting repair. The MC/NMC rate steady states can be found similarly to be p p p p p p 2 2 1 1 2 1 1 1 1 - + - - + L N M M M M O Q P P P P ( ) . For example, let’s let p1 =.95 and p2 =.8 . This means that only 5% of the working vehicles break each day, and 80% of the broken vehicles are repaired. We set T = 58 . Substitution gives us a steady state of 54 5882 34118 . . L N M O Q P , and an eventual steady state OR of 94.1176%. Of course, we can’t have 54.5882 tanks. That is the expected value for the number operational, which is an average. Just as the average of 1 and 2 is 1.5, which is not an integer, so can the average number of tanks operational also not be an integer. We can verify these calculations by constructing a spreadsheet model of the tank maintenance status. Let’s start with M = 50 and N = 4. Figure 2 is a picture of how the system on average would behave
55545352515004681021214161820Figure 2Let'slook a little closer at the steady state for M, which we said was equal toT p2M..P2-p, +1If wewanttoimprovethe expected steadystatefor M,shouldwetryto increasePr, P2, or some combination of the two? One way to gain insights into thisquestion is to look at the gradient of steady state for M with respect to p1 and p2ThegradientofM isgivenby:LaMaMT(1-p)T p2VMseeadystateNopap,UN(p, - p, + 1)2'(p2 - p + 1)2Weknowfrom calculus that the direction of greatest increase in M is in thedirection of the gradient. That means to obtain the greatest increase in M for afixedsizechangeinthevaluesofpandp2,thechangeinpishouldbeproportionaltop2,andthechangeinp2shouldbeproportionalto1-p1.Doesthismakesense?If we are repairing vehicles quickly already,then itmakes sense to work on notbreakingtheminsteadofworkingon improving ourrepairtime.Thatis whythechangeinpisproportionaltop2.Asp2increases,weworkmoreonp1.172
172 50 51 52 53 54 55 0 2 4 6 8 10 12 14 16 18 20 Figure 2 Let’s look a little closer at the steady state for M, which we said was equal to M T p p p steady state = - + 2 2 1 1 . If we want to improve the expected steady state for M, should we try to increase p1 , p2 , or some combination of the two? One way to gain insights into this question is to look at the gradient of steady state for M with respect to p1 and p2. The gradient of M is given by: Ñ = L N M O Q P = - + - - + L N M O Q M P M p M p T p p p T p p p steady state ¶ ¶ ¶ 1 ¶ 2 2 2 1 2 1 2 1 2 1 1 1 , ( ) , ( ) ( ) . We know from calculus that the direction of greatest increase in M is in the direction of the gradient. That means to obtain the greatest increase in M for a fixed size change in the values of p1 and p2, the change in p1 should be proportional to p2, and the change in p2 should be proportional to 1-p1. Does this make sense? If we are repairing vehicles quickly already, then it makes sense to work on not breaking them instead of working on improving our repair time. That is why the change in p1 is proportional to p2. As p2 increases, we work more on p1
Ifwearebreakingvehiclesfrequently,itmakessensetoimproveourrepaircapability.(1-p1)is thefraction of vehicles expected to“break"each timeperiod.Asthat increases,so shouldp2,accordingtothegradient.Noticethat unlessp2is zero, orp,isone,we should work on improving bothpercentages. The gradient tells us how to allocate our effort between them.There are at least two shortcomings of this model.First, it is deterministic:weonlyworkwiththeexpectednumberofvehiclesineachcategory,andthisignores therandomness of the true behavior,as well as the integer nature of theobjects.Wecan'thave54.5882tanksmissioncapable:wecanonlyhave54or55.Wedon'tknowfromthismodelhowmuchvariationtoexpectaroundtheaverage.Thesecondshortcomingisthatit isverysimple.Itdoesn'trecognizethatvehiclesbreakatdifferentratesdependinguponwhetheroneisdrivingtheminthe field or not. Vehicles are repaired at different rates depending upon theavailabilityofsparepartsandmechanics,anduponhowmanyothervehiclesarewaitingtoberepaired.Theavailabilityofpartsdependsontheprioritycodeofthe unit.If the unit level maintenance can not repair the vehicle,it is sent tohigherlevelmaintenance,whichhasdifferentrepairrates.Abettermodelwould incorporatethese additional featuresHowever, this simple model is useful.It allows us to gain understandingaboutoursystem,whichisthehallmarkofaneffectivemodel.Binomial EquationItispossibleto improvethemodel intheprevioussectionbyconsideringastochasticmodel.Thiswillallowustonotonlyunderstandtheaveragebehaviorof our model, butalso howmuch thenumber of Mctanksvaries around thataverage.There isa significantdifferencebetween54.5882(plus orminus.1),and 54.5882(plus orminus 10).Wewill usethebinomial distribution forourmodels.Let's considerthe number oftanksthat are mission capable each day.Thatnumberwillbethesumofthosetanksthatweremissioncapablethepreviousdayandstayed inthat state,plusthe number oftanks that were repaired theprevious day.Thiscanbemodeled asthesumoftwobinomiallydistributedrandomvariables.LetM,bethenumberofmissioncapabletankstheprecedingperiod,andTbethetotal number oftanks.ThenM+=Bin(pr, M)+Bin(p2,T-M)173
173 If we are breaking vehicles frequently, it makes sense to improve our repair capability. (1-p1 ) is the fraction of vehicles expected to “break” each time period. As that increases, so should p2, according to the gradient. Notice that unless p2 is zero, or p1 is one, we should work on improving both percentages. The gradient tells us how to allocate our effort between them. There are at least two shortcomings of this model. First, it is deterministic: we only work with the expected number of vehicles in each category, and this ignores the randomness of the true behavior, as well as the integer nature of the objects. We can’t have 54.5882 tanks mission capable: we can only have 54 or 55. We don’t know from this model how much variation to expect around the average. The second shortcoming is that it is very simple. It doesn’t recognize that vehicles break at different rates depending upon whether one is driving them in the field or not. Vehicles are repaired at different rates depending upon the availability of spare parts and mechanics, and upon how many other vehicles are waiting to be repaired. The availability of parts depends on the priority code of the unit. If the unit level maintenance can not repair the vehicle, it is sent to higher level maintenance, which has different repair rates. A better model would incorporate these additional features. However, this simple model is useful. It allows us to gain understanding about our system, which is the hallmark of an effective model. Binomial Equation It is possible to improve the model in the previous section by considering a stochastic model. This will allow us to not only understand the average behavior of our model, but also how much the number of MC tanks varies around that average. There is a significant difference between 54.5882 (plus or minus .1), and 54.5882 (plus or minus 10). We will use the binomial distribution for our models. Let’s consider the number of tanks that are mission capable each day. That number will be the sum of those tanks that were mission capable the previous day and stayed in that state, plus the number of tanks that were repaired the previous day. This can be modeled as the sum of two binomially distributed random variables. Let Mi be the number of mission capable tanks the preceding period, and T be the total number of tanks. Then Mi+ = Bin p Mi Bin p T Mi + - 1 1 2 ( , ) ( , )